# Variations on confidence intervals

The radius of a confidence interval, z_(*alpha*/2) × *sigma*/(n^.5), depends on the level of confidence (1-*alpha*), and the sample size (n). This suggests three questions we may ask related to confidence intervals: The most common problem is to construct a confidence interval, i.e., given 1-*alpha* and r, find the radius of the confidence interval. This is what we did in the previous section.

Another problem asks how large n must be to get a confidence interval of a specified radius. (Recall that the larger n is, the smaller the radius of the confidence interval is.
Example: If *sigma* = 15, how large must n be to get a 90% confidence interval with radius less than or equal to .5?
90% confidence provides *alpha*/2 = .05; looking up .95 in the body of the normal table gives z_.05=1.65. Therefore we want to solve .5 > 1.65 × 15/(n^.5) which is equivalent to n^.5 > 1.65 × 15/.5 = 49.5. Squaring both sides provides n must be greater than 2450.25, or at least 2451 since n is an integer.

Sometimes, one may have collected the data, and wonder how confident he can be that the mean is within a given distance of x-bar.
Example: If x-bar = 39 based on a sample of size 60, and *sigma* = 20, how confident are you that the interval (35, 43) contains µ?
Since 4 is the radius of the confidence interval, 4=z_(*alpha*/2) × *sigma*/(n^.5). Since *sigma*=20 and n=60, this becomes 4=z_(*alpha*/2) × 20/(60^.5), which can be solved for z_(*alpha*/2) = 1.5492. From the normal table we find *alpha*/2=.0606, hence the level of confidence, (1-*alpha*), is .8788.

Competencies: If the standard deviation of your population is 13, how large must the sample size be to get a 95% confidence interval with a margin of error (radius) less than or equal to 2?
If the standard deviation of your population is 12 and based on a sample of size 60 you get the confidence interval (17, 22), what is the level of confidence of that interval (recall that x-bar is in the middle of the confidence interval).